Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm reading the Atiyah-MacDonald book on Commutative Algebra. At the beginning of the module chapter on page 17, they make an example which I don't understand. Example 5) is:

$G$ = finite group, $A = k[G] =$ group-algebra of $G$ over the field $k$ (thus $A$ is not commutative, unless $G$ is). Then $A$-module $= k$-representation of $G$.

I'm far away from understanding how $k[G]$-module = $k$-representation of $G$. I try to unwind the definitions as follows:

After reading the Wikipedia articles on monoid rings and group rings, I come to the conclusion that a group algebra is a triple $(k[G], +, \star)$, where \begin{align} k[G] = \{ \phi: G \rightarrow R \mid \{g \in G \mid \phi(g) \neq 0 \}\ \text{finite} \}, \tag{1} \end{align} and where + is pointwise adition and $\star$ is the convolution \begin{align} (\phi \star \psi)(g) = \sum\limits_{kl = g} \phi(k) \psi(l). \end{align} According to the definition of a module on the same page (17) in Atiyah-MacDonald, a $k[G]$-module is a pair $(M, \mu)$, where $M$ is an (additive) Abelian group and $\mu$ is a map \begin{align} \begin{array}{rccl} \mu: &k[G] \times M &\rightarrow &M \\ &(\phi, x) &\mapsto &\phi x := \mu(\phi, x) \end{array} \end{align} such that \begin{align} \phi(x + y) &= \phi x + \phi y, \\ (\phi + \psi) x &= \phi x + \psi x, \\ (\phi \psi) x &= \phi(\psi x), \\ 1 x &= x. \end{align}

Question 1: How is specifying a $k[G]$-module equivalent to specifying a $k$-representation of $G$?

Question 2: Why is finiteness of $G$ needed? Does it have something to do with the finiteness in definition (1) of $k[G]$?

share|improve this question
    
It's easier to work with the definition of $k[G]$ as the formal sums $\sum_{g \in G} a_g e_g$ where $a_g \in k$ and $\{e_g\}_{g \in G}$ is a basis for $k[G]$ as a $k$-vector space. Multiplication in $k[G]$ is defined by $e_g e_h = e_{gh}$ and extended linearly. Try to associate to each representation $k[G] \rightarrow \text{End(V)}$ (the right-side is the algebra of endomorphisms of a vector space $V$; concretely it is just the ring of $n \times n$ matrices when you equip $V$ with a basis) a group representation $G \rightarrow GL(V)$. You'll want to use that $e_g e_{g^{-1}} = 1$ in $k[G]$. –  Michael Joyce Apr 20 '13 at 11:28

1 Answer 1

up vote 8 down vote accepted

This equivalence is a very formal matter, and the finiteness of $G$ (nor any properties of the field $k$, nor that fact that $G$ is a group rather than just a monoid) does not play a role in it. The basic information is given in both cases by a a $k$-vector space $V$ equipped with a "multiplication" map $m:G\times V\to V$ that must satisfy

  • associativity with respect to the first argument: $m(g_1,m(g_2,v))=m(g_1g_2,v)$ for all $g_1,g_2\in G$ and $v\in V$

  • $k$-linearity with respect to the second argument: $m(g,\lambda v_1+\mu v_2)=\lambda m(g,v_1)+\mu m(g,v_2)$ for all $g\in G$, $v_1,v_2\in V$ and $\lambda,\mu\in k$.

If in this description you fix individual group elements $g$, then the second point says that $v\mapsto m(g,v)$ is a $k$-linear map $\rho_g:V\to V$, and then the first condition says that $\def\End{\operatorname{End}}\rho:G\to\End_k(V)$ is a monoid homomorphism, and by restriction of codomian a group homomorphism $G\to GL_k(V)$ is $G$ is a group, in other words this precisely describes a $k$-representation of $G$.

If one the other hand you start with $V$ as Abelian group, then you've got two additional structures: the $k$-vector space structure that defines scalar multiplications by $\lambda\in k$, and the multiplications $\rho_g$, which are compatible with the vector space structure (they are $k$-linear) and define a monoid action (the first point above). These are precisely the conditions needed to be able to define multiplication by formal $k$-linear combinations of elements of $G$ by the rule $$ \mu(\sum_i \lambda_i g_i), v)=\sum_i \lambda_i m(g,v) $$ whenever the summation is finite (infinite linear combinations make no sense, unless all but finitely many coefficients are zero), and so that $\mu$ satisfies $\mu(\phi\psi,v)=\mu(\phi,\mu(\psi,v))$ where multiplication of formal linear combinations of group elements is defined by extending the group multiplication by bilinearity: $$ (\sum_i \lambda_i g_i)(\sum_j \kappa_j g_j)=\sum_{i,j}\lambda_i\kappa_j g_ig_j $$ But the collection of formal linear combinations with that multiplication (and the obvious $k$-vector space structure) is just the algebra $k[G]$, and we've just required that $\mu$ be a $k[G]$ module (the $k$-vector space axioms for $V$ are just the set of these requirements limited to scalars in the subalgebra $k=k[e]$ of $k[G]$.)

Just a final word; as you see the finiteness of the number of elements of $G$ that have nonzero coefficient in an element of $k[G]$ is just something that is required to make the formal expressions used have sense, and in particular to have a well defined multiplication on $k[G]$. They do not restrict $G$ itself in any way (indeed they are precisely there to be able to do this construction equally well when $G$ is infinite as when it is finite).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.